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Vn Method

Let r th term...

Question

Let $r^{th}$ term of a series be given by $T_{r} = \frac{r}{1 - 3 r^{2} + r^{4}} .$ Then $- 2 \infty \sum r = 1 T_{r}$ is

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Solution

$T_{r} = \frac{r}{1 - 3 r^{2} + r^{4}} = \frac{r}{(r^{2} - 1)^{2} - r^{2}} = \frac{1}{2} [\frac{1}{r^{2} - 1 - r} - \frac{1}{r^{2} - 1 + r}]$
So, $n \sum r = 1 T_{r}$
$= \frac{1}{2} [- 1 - 1] + \frac{1}{2} [1 - \frac{1}{5}] + \frac{1}{2} [\frac{1}{5} - \frac{1}{11}] + \frac{1}{2} [\frac{1}{11} - \frac{1}{19}] + ⋮ ⋮ + \frac{1}{2} [\frac{1}{n^{2} - 1 - n} - \frac{1}{n^{2} - 1 + n}] S_{n} = \frac{1}{2} [- 1 - \frac{1}{n^{2} - 1 + n}]$
For very large value of denominator, the fraction will be $≅ 0$
$∴ S_{\infty} = - \frac{1}{2}$

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